Problem: A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(-12,-5t\right)$ for time $t\geq 0$. At $t=3$, the particle is at the point $(-1,-2)$. What is the particle's position at $t=5$ ? $($
To find the particle's position at $t=5$, we need to find its horizontal displacement $\Delta x$ and its vertical displacement $\Delta y$, and add those to its initial position $(-1,-2)$ : $\text{Position at }t=5\text{: }(-1+\Delta x,-2+\Delta y)$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between $t=3$ and $t=5$ : $\Delta x=\int_{3}^{5} -12\,dt=-24$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between $t=3$ and $t=5$ : $\Delta y=\int_{3}^{5} -5t\,dt=-40$ Now we can find the particle's position: $\begin{aligned} &\phantom{=}(-1+\Delta x,-2+\Delta y) \\\\ &=\left(-1+\left(-24\right),-2+\left(-40\right)\right) \\\\ &=(-25,-42) \end{aligned}$ In conclusion, particle's position at $t=5$ is $(-25,-42)$.